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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T07:25:01+00:00 2026-05-13T07:25:01+00:00

In the ocaml language specification, there’s a short section: poly-typexpr ::= typexpr | {

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In the ocaml language specification, there’s a short section:

poly-typexpr ::= typexpr
               | { ' ident }+ . typexpr

There’s no explanation in the text, and the only instance of poly-typexpr is in defining a method type:

method-type ::= method-name : poly-typexpr

What does this allow me to do?

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  1. Editorial Team
    Editorial Team
    2026-05-13T07:25:01+00:00Added an answer on May 13, 2026 at 7:25 am

    poly-typexpr is also allowed as the type of a record field (see Section 6.8.1). These are commonly called “existential types,” though there is some debate on that point. Using a polymorphic type in this way changes the scope of the type variable. For example, compare the types:

    type 'a t = { f : 'a -> int; }
    type u = { g : 'a. 'a -> int; }
    

    t is really a family of types, one for each possible value of 'a. Each value of type 'a t must have a field f with the type 'a -> int. For example:

    # let x = { f = fun i -> i+1; } ;;
    val x : int t = {f = <fun>}
    # let y = { f = String.length; } ;;
    val y : string t = {f = <fun>}
    

    In comparison, u is a single type. Each value of type u must have a field g with the type 'a -> int for any 'a. For example:

    # let z = { g = fun _ -> 0; } ;;
    val z : u = {g = <fun>}
    

    Note here that g doesn’t depend on the type of its input at all; if it did, it wouldn’t have the type 'a. 'a -> int. For example:

    # let x2 = { g = fun i -> i+1; } ;;
    This field value has type int -> int which is less general than 'a. 'a -> int
    
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